Gibbs Free Energy Video Lessons

Concept

# Problem: Given:P4(s) + 5O2(g)⟶P4O10(s)                  ΔG° 298 = −2697.0 kJ/mol2H2(g) + O2(g)⟶2H2O(g)                  ΔG° 298 = −457.18 kJ/mol6H2O(g) + P4O10(s)⟶4H3 PO4(l)      ΔG°298 = −428.66 kJ/molDetermine the standard free energy of formation, ΔG°f, for phosphoric acid. How does your calculated result compare to the value in Appendix G? Explain.

###### FREE Expert Solution

Part A

Reaction 1 no change

P4 (s) + 5 O2 (g) P4O10 (s)                    ΔG° 298 = −2697.0 kJ/mol

Reaction 2  multiply by 3

6 H2(g) + 3 O2(g) 6 H2O(g)                  ΔG° 298 = 3(−457.18 kJ/mol) = -1371.54 kJ/mol

Reaction 3  no change

6H2O(g) + P4O10(s)4H3 PO4(l)             ΔG°298 = −428.66 kJ/mol

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###### Problem Details

Given:
P4(s) + 5O2(g)⟶P4O10(s)                  ΔG° 298 = −2697.0 kJ/mol
2H2(g) + O2(g)⟶2H2O(g)                  ΔG° 298 = −457.18 kJ/mol
6H2O(g) + P4O10(s)⟶4H3 PO4(l)      ΔG°298 = −428.66 kJ/mol

Determine the standard free energy of formation, ΔG°f, for phosphoric acid. How does your calculated result compare to the value in Appendix G? Explain.